Open In App

Find sum of the series 1-2+3-4+5-6+7…….

Last Updated : 17 Nov, 2022
Improve
Improve
Like Article
Like
Save
Share
Report

Given a number N. The task is to find the sum of the below series up to nth term.
 

1- 2 + 3 – 4 + 5 – 6 +….

Examples
 

Input : N = 8
Output : -4

Input : N = 10001
Output : 5001

 

Approach: If we observe carefully, we can see that the sum of the above series follows a pattern of alternating positive and negative integers starting from 1 to N as shown below:
 

N   = 1, 2, 3, 4, 5, 6, 7 ......
Sum = 1, -1, 2, -2, 3, -3, 4 ......

Hence, from the above pattern, we can conclude that: 
 

  • when n is odd => sum = (n+1)/2
  • when n is even => sum = (-1)*n/2

Below is the implementation of the above approach:
 

C++




// C++ program to find the sum of
// series 1 - 2 + 3 - 4 +......
 
#include <iostream>
using namespace std;
 
// Function to calculate sum
int solve_sum(int n)
{
    // when n is odd
    if (n % 2 == 1)
        return (n + 1) / 2;
 
    // when n is not odd
    return -n / 2;
}
 
// Driver code
int main()
{
    int n = 8;
     
    cout << solve_sum(n);
     
    return 0;
}


Java




// Java program to find sum of
// first n terms of the given series
import java.util.*;
 
class GFG
{
static int calculateSum(int n)
{
    // when n is odd
    if (n % 2 == 1)
        return (n + 1) / 2;
 
    // when n is not odd
    return -n / 2;
}
 
// Driver code
public static void main(String ar[])
{
 
// no. of terms to find the sum
int n = 8;
System.out.println(calculateSum(n));
}
}


Python 3




# Python program to find the sum of
# series 1 - 2 + 3 - 4 +......
 
# Function to calculate sum
def solve_sum(n):
    # when n is odd
    if(n % 2 == 1):
        return (n + 1)/2
 
    # when n is not odd
    return -n / 2
 
# Driver code
n = 8
 
print(int(solve_sum(n)))


C#




// C# program to find sum of
// first n terms of the given series
using System;
 
class GFG
{
static int calculateSum(int n)
{
    // when n is odd
    if (n % 2 == 1)
        return (n + 1) / 2;
 
    // when n is not odd
    return -n / 2;
}
 
// Driver code
public static void Main()
{
 
    // no. of terms to find the sum
    int n = 8;
    Console.WriteLine(calculateSum(n));
}
}
 
// This code is contributed
// by inder_verma


PHP




<?php
// PHP program to find the sum of
// series 1 - 2 + 3 - 4 +......
 
// Function to calculate sum
function solve_sum($n)
{
    // when n is odd
    if ($n % 2 == 1)
        return ($n + 1) / 2;
 
    // when n is not odd
    return -$n / 2;
}
 
// Driver code
$n = 8;
 
echo solve_sum($n);
 
// This code is contributed
// by inder_verma
?>


Javascript




<script>
 
// javascript program to find sum of
// first n terms of the given series
 
 
 
function calculateSum(n)
{
    // when n is odd
    if (n % 2 == 1)
        return (n + 1) / 2;
 
    // when n is not odd
    return -n / 2;
}
 
// Driver code
 
// no. of terms to find the sum
var n = 8;
document.write(calculateSum(n));
 
 
// This code contributed by shikhasingrajput
 
</script>


Output: 

-4

 

Time Complexity: O(1)

Auxiliary Space: O(1) , since no extra space has been taken.



Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads